Recent content by Katte

Oh. Wow! Well, thank you for all the help! I realize now the answer was pretty simple, all things considered. Thanks for steering me in the right direction! =)

Okay, let's see if I got this. (R^3)(density_ave)=(r^3)(density_core)+(R^3-r^3)(density_mantle) ((2.4397x10^6m)^3)(5.43x10^3kg/m^3)=(r^3)(7,800kg/m^3)+((2.4397x10^6m)^3-r^3)(2,600kg/m^3) Which calculates to (rounded for reading's sake): (7.89x10^22)=(7,800r^3)+(3.78x10^22)-(2,600r^3)...

Total Mass = v x d = (6.081 x 10^19 m3)(5.43 x 10^3 kg/m3) Mantle Mas = v x d = (4/3(pi)R-r)3 x d = (4/3 (pi) 2.439x10^6m - r)3 x (2,600 kg/m3) Core Mass =v x d =(4/3(pi)r)3 x d = (4/3 (pi) r)3 x (7,800 kg/m3)

Would I possibly use the total radius to determine Mercury's moment of inertia to calculate core size? I've just read about it and unsure of the process.

So I have calculated that the mass of Mercury is 3.303 x 10^23 kg, and the volume of Mercury is 6.08 x 10^19 m3, But I'm unsure of how to calculate the volume or mass of the mantle, because wouldn't I need the volume or mass of the core to do so?

Homework Statement After determining the average density of Mercury as 5.43 x 10^3(kg/m3), you may calculate the size of Mercury's core. Assume Mercury consists of a core and a mantle+crust, with core composition being similar to Earth (7,800(kg/m3)). From space missions and studying...